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For a lower triangular matrix, the inverse of itself should be easy to find because that's the idea of the LU decomposition, am I right? For many of the lower or upper triangular matrices, often I could just flip the signs to get its inverse. For eg: $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -1.5 & 0 & 1 \end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 1.5 & 0 & 1 \end{bmatrix}$$ I just flipped from -1.5 to 1.5 and I got the inverse.

But this apparently doesn't work all the time. Say in this matrix: $$\begin{bmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 3.5 & -2.5 & 1 \end{bmatrix}^{-1}\neq \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ -3.5 & 2.5 & 1 \end{bmatrix}$$ By flipping the signs, the inverse is wrong. But if I go through the whole tedious step of gauss-jordan elimination, I would get its correct inverse like this: $\begin{bmatrix} 1 & 0 & 0\\ -2 & 1 & 0\\ 3.5 & -2.5 & 1 \end{bmatrix}^{-1}= \begin{bmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ 1.5 & 2.5 & 1 \end{bmatrix}$ And it looks like some entries could just flip its signs but not for others.

Then this is kind of weird because I thought the whole idea of getting the lower and upper triangular matrices is to avoid the need to go through the tedious process of gauss-jordan elimination and can get the inverse quickly by flipping signs? Maybe I have missed something out here. How should I get an inverse of a lower or an upper matrix quickly?

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    $\begingroup$ you can only flip the signs for atomic triangular matrices. The first one is atomic, the second one is not. $\endgroup$ – Nana Jun 25 '11 at 6:56
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Ziyuang's answer handles the cases, where $N^2=0$, but it can be generalized as follows. A triangular $n\times n$ matrix $T$ with 1s on the diagonal can be written in the form $T=I+N$. Here $N$ is the strictly triangular part (with zeros on the diagonal), and it always satisfies the relation $N^{n}=0$. Therefore we can use the polynomial factorization $1-x^n=(1-x)(1+x+x^2+\cdots +x^{n-1})$ with $x=-N$ to get the matrix relation $$ (I+N)(I-N+N^2-N^3+\cdot+(-1)^{n-1}N^{n-1})=(I-N^n)=I $$ telling us that $(I+N)^{-1}=I+\sum_{k=1}^{n-1}(-1)^kN^k$.

Yet another way of looking at this is to notice that it also is an instance of a geometric series $1+q+q^2+q^3+\cdots =1/(1-q)$ with $q=-N$. The series converges for the unusual reason that powers of $q$ are all zero from some point on. The same formula can be used to good effect elsewhere in algebra, too. For example, in a residue class ring like $\mathbf{Z}/2^n\mathbf{Z}$ all the even numbers are nilpotent, so computing the modular inverse of an odd number can be done with this formula.

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In case of a lower triangular matrix with arbitrary non-zero diagonal members, you may just need to change it in to: $T = D(I+N)$ where $D$ is a diagonal matrix and $N$ is again an strictly lower diagonal matrix. Apparently, all said about inverse in previous comments will be the same.

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Under the assumption of diagonal entries being 1, what you mean is $(I+N)^{-1}=(I-N)$, where $N$ is a nilpotent matrix, yielding $N^2=0$.

If diagonal entries are not all 1's (and none is 0), denote the matrix with only those diagonal entries as $D$, it will be reduced to $N^2=ND^{-1}-DN=D^{-1}N-ND$. Waiting for further simplification.

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Computing the inverse misses the whole point of factorizing into triangular matrices. If you have a triangular matrix, you should almost never need to compute the inverse, because solving triangular systems can be done quickly by back/forward-substitution without ever inverting the matrix.

For example, if you have a lower-triangular matrix $L$, and you see an expression like $L^{-1} y$, you should almost never actually compute $L^{-1}$ and then multiply it by $y$. Instead, you would solve $Lx = y$ by forward-substitution, obtaining $x=L^{-1}y$.

More quantitatively, if you have an $m \times m$ upper/lower triangular matrix $T$, then you can solve $Tx=y$ by back/forward-substitution in $\Theta(m^2)$ operations, whereas computing $T^{-1}$ for a general triangular matrix requires $\Theta(m^3)$ operations. (In general, when you see $A^{-1}y$ and you want to compute it, you should read it as "solve $Ax=y$ by the best available method for $A$"; computational scientists almost never explicitly compute inverse matrices.)

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protected by Zev Chonoles Mar 3 '16 at 6:06

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